If y= ^{-1}≤ft[{x-{1-x²}}{x+{1-x²}}], then d y/d x=

Correct Answer is : 11−x2{1}{{1-x²}}1−x21

Correct Answer is : 11−x2{1}{{1-x²}}1−x21

Test Index

MHTCET 16 Apr 2026 Shift 1 Paper

Question : 116 of 150

Marks: +1, -0

y=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right]

\frac{d y}{d x}=

Solution:

Concept:

Use the substitution

x = \sin\theta

to simplify the expression inside

\tan^{-1}

and then differentiate.

Explanation:

Let

x = \sin\theta

, where

-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

Then

\sqrt{1-x^{2}} = \cos\theta

Substitute into the given expression:

\frac{x - \sqrt{1-x^{2}}}{x + \sqrt{1-x^{2}}} = \frac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta}.

Using the identity

\tan\left(\theta - \frac{\pi}{4}\right) = \frac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta}

, we get:

y = \tan^{-1}\left[\tan\left(\theta - \frac{\pi}{4}\right)\right].

y = \theta - \frac{\pi}{4}

up to a constant.

A constant term disappears on differentiation, hence:

\frac{dy}{dx} = \frac{d\theta}{dx}.

Since

x = \sin\theta

, we have:

\frac{dx}{d\theta} = \cos\theta = \sqrt{1-x^{2}}.

Therefore:

\frac{dy}{dx} = \frac{1}{\frac{dx}{d\theta}} = \frac{1}{\sqrt{1-x^{2}}}.

Answer:

The correct option is B:

\frac{1}{\sqrt{1-x^{2}}}

Go to Question:

Prev Question

Next Question